A ring of radius R is charged uniformly with a charge +Q . The electric field at any point on its axis at a distance r from the circumference of the ring will be
A
KQr
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B
KQr2
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C
KQr3(r2−R2)12
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D
KQrR3
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Solution
The correct option is CKQr3(r2−R2)12
We know that, electric field at any point on the axis is given by, E=KQx(R2+x2)32
and from diagram, we get x=√r2−R2
Therefore, E=KQ√r2−R2(R2+r2−R2)32=KQr3(r2−R2)12